Vector addition and subtraction
Vector addition and subtraction problems can be solved by triangle and parallelogram law.
Triangle law of vector addition:
Statement: If two vectors are represented both in magnitude and direction by two sides of a triangle taken in the same order, then the resultant of these vectors is represented both in magnitude and direction by the third side of the triangle taken in the opposite order.
Prove: Let us consider 
and 
acting simultaneously on a particle be represented both in magnitude and direction by two sides OA and AC of 
OAC taken in the same order. 
is the resultant of and is represented by side OC. So, = +
CN is the perpendicular on ON and θ is the angle between and .
In right-angled ONC, OC2 = ON2 +NC2
Or, R2 = (OA + AN)2 + NC2
Or, R2 = OA2 + AN2 + 2OA.AN + NC2
Or, R2 = OA2 + AN2 + 2OA.ACcos
+ NC2
R2 = OA2 + AC2 + 2OA.ACcos
[From right-angled ANC, 
= 
so, AN = AC and AN2+ NC2 = AC2]
R2 = P2 + Q2 + 2PQ
Let us consider 
creates angle 
with 
. Then from right-angled ONC, 
= 
= 
= 
Or, = 
[From right-angled ANC, 
= 
].
Parallelogram law of vector addition:
Statement: If two vectors are represented both in magnitude and direction by two adjacent sides of a parallelogram drawn from a point, then their resultant will be represented both in magnitude and direction by the diagonal of the parallelogram drawn from that point.
Prove: Same as triangle law.
Vector addition follows commutative law: When two vectors are added, the sum is independent of the order of the vector addition.
Vector addition follows associative law: For addition of three or more vectors the sum is independent of the way in which the individual vectors are grouped together.
Special case:
- If two vectors
and 
are in same direction, then the angle between them is 00 and the resultant is maximum. Hence, 
= P2 + Q2 + 2PQcos
= (P + Q)2
= P + Q.
- If two unequal vectors are in opposite direction (i.e. anti parallel) then the angle between them is 1800 and the resultant is minimum. Hence,
= P2 + Q2 + 2PQcos
= (P – Q)2
= P – Q.
When two vectors of equal magnitude are in opposite direction then the resultant is zero this is known as null vector.
- If two vectors and are perpendicular to each other, then, the resultant is R2 = P2 + Q2 + 2PQcos
R = 
.
Subtraction of vectors:
Let us consider and creates angle θ. To subtract from we can write + (- ) [where – is the opposite vector of ]. The angle between and – is (180 – θ) and is the resultant of and – .
The magnitudes of is R2 = P2 + Q2+2P.Q
Or, R2 = P2 +Q2– 2PQ
If creates angle 
with then tan = 
tan = 
.
Example: 1. The sum of the magnitude of two forces act at a point is 8 N. If their resultant is normal to the smaller force and has magnitude of 4 N, then find the magnitude of forces.
Let us consider, the magnitude of smaller force is P and the other force is (8 – P). As the resultant is normal to P then, (8 – P)2 = 42 + P2 or, 64 – 16P + P2 = 16 + P2 or, P = 
= 3 So, magnitude of forces are 3 and 5.
2. Two vectors are such a way that| 
+ 
| = | – |, then calculate the angle between the two vectors. [A 
0, B 0]
Let the angle between and is 
as, | + | = | – | then, A2 +B2 + 2ABcos = A2 +B2 – 2ABcos
Or, 4ABcos = 0
So, θ = 900.
3. The resultant of two forces 
and 
is 
. If is reversed then the resultant is 
, then prove that R2 + S2 = 2(F12 + F22).
From question, = + so, R2 = F12 + F22 +2
cos —-(1)
And = – so, Q2 = F12 + F22 -2cos ——(2)
From equation (1) and equation (2) we get, R2 + Q2 = 2(F12 + F22).
4. The resultant of and is . If the magnitude of is doubled, the new resultant becomes perpendicular to . Find the magnitude of .
From question, R2 = P2+Q2+2PQcos —(1) and tan 
= 
where θ is the angle between and and is the angle between and . When is doubled resultant is 
.
So, R12 = P2+4Q2+4PQcos —(2) as R1 is perpendicular to P, then R12 + P2 = 4Q2 —(3).
From equation (2) and equation (3) P2 + 2PQcos = 0 —-(4). From equation (1) and equation (4) R2 = Q2. So, R = Q.
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